If $H,K\leq G$ with $[G:K]=n,[G:H]=m$, then $[G:H\cap K]\geq \text{lcm}(m,n)$?

Question

I have the following question:

If $H,K\leq G$ with $[G:K]=n,[G:H]=m$, then show that $[G:H\cap K]\geq \text{lcm}(m,n)$ and the equality holds when $gcd(m,n)=1$.

Attempt

I have shown that $[G:H\cap K]=n[K:H\cap K],m[H:H\cap K]$. But I don't know how to go on.

This doesn't make any sense: $[G:H \cap K]= n|H \cap K|,m|H \cap K|$. Please edit to say what you mean. — fred goodman, Feb 10 '18 at 15:10
You have not fixed your attempt yet. Please do. I have mistakenly pressed the reopen button. — , Feb 11 '18 at 14:17
Looks very much like this one: https://math.stackexchange.com/q/86086/9464 — , Feb 11 '18 at 14:18

score 1 · Accepted Answer · answered Feb 10 '18 at 15:49

We know that $H \cap K$ is a subgroup of both $K$ and $H$. Therefore we have that

$$[G:H \cap K] = [G:H][H:H \cap K] \implies m = [G:H] \big |[G:H \cap K]$$

Similarly:

$$[G:H \cap K] = [G:K][K:H \cap K] \implies n = [G:K] \big |[G:H \cap K]$$

So therefore we have that $[G:H \cap K]$ is a common multiple of $m,n$ and hence trivially $[G:H \cap K] \ge \text{lcm}(m,n)$.

For the second part use the identity $\text{lcm}(m,n) \cdot \gcd(m,n) = mn$ and the well-known inequality $[G:H \cap K] \le mn$.

score 1 · Answer 2 · answered Feb 10 '18 at 15:49

$|G:(H\cap K)|=|G:H||H:(H\cap K)|=|G:K||K:(H\cap K)|$

Therefore $|G:(H\cap K)|$ is divisible by $|G:H|$ and by $|G:K|$ and therefore is divisible by $l.c.m.(|G:H|, |G:K|)$, i.e. $|G:(H\cap K)|\leq l.c.m.(|G:H|, |G:K|)$

On the other side, since $x(H\cap K)=xH\cap xK$, we have $|G:(H\cap K)|\leq |G:K||G:H|$. And if $|G:K|$ and $|G:H|$ are coprime, then $l.c.m.(|G:H|, |G:K|) =|G:H||G:K|$, so, $|G:(H\cap K)|= l.c.m(|G:K|,|G:H|)$.

If $H,K\leq G$ with $[G:K]=n,[G:H]=m$, then $[G:H\cap K]\geq \text{lcm}(m,n)$?

2 Answers2